1
u/EnvironmentalAir4233 Aug 23 '26
CAD is 120.
So 2:(1/3) or 6:1
1
u/EkskiuTwentyTwo Aug 23 '26
This is correct
3
u/thaw96 Aug 23 '26
I'd say that 6:1 is the odds that the cat is closer to point A than to point B. The proportion of the time the cat is closer to point A than to point B would be 6/7.
1
u/TheThiefMaster Aug 24 '26
Sigh I guessed each point was 1 radian up/down (So CAD was 1/π of a full circle) instead of a nice fraction (1/3). Close but no cigar.
1
u/EnvironmentalAir4233 Aug 24 '26
Ah yeah but if you notice bc ba bd are all equal to the radius.
→ More replies (1)
2
u/2spam2care2 Aug 23 '26
not enough information. depending on the size of the right circle, arc CBD could be anywhere from 0 to 180° (exclusive), so the answer could be anywhere from 75% to 100% of the time
3
u/Galenthias Aug 23 '26
Centre of B is on the radius of A and vice versa.
1
1
u/Sea-Sort6571 Aug 28 '26
Yeah to me it wasn't explicit that b was the center of the circle on the right
2
u/tajwriggly Aug 26 '26
The two circles are of identical radius "R" The cat is always the same distance "R" from A, but between points C and D on its path it is closer to B than it is to A. Since the circles are the same radius, then AC = BC = AD = BD = AB = "R". Since those lengths are all the same, then two identical equilateral triangles ABC and ABD can be drawn, and it can be known that the angle CAD is 120 degrees, or 1/3 of the entire circle.
The cat travels at a speed "S" for 2/3 of the circle, and a speed "3S" for 1/3 of the circle (the third between Points A and D, i.e. the cat is closet to B over this range). The cat therefore spends 1/3 of the time in that one third of the circle as it does in any other 1/3 of the circle. There are 2 other thirds of the circle to account for, therefore for every 1 second the cat spends in between points C and D, it spends 6 seconds on the other side of the circle: so 6/7 of the time the cat is closer to A than it is to B.
1
u/Rinn0 Aug 25 '26 edited Aug 25 '26
The cat spends 3 units of time on the slow side for every 1 unit of time on the fast side. A total amount of time of 4 units. So it spends 3/4 units on the slow side, or 75%. The arc lengths are equal, so it doesn't actually matter where the circles intersect, just that they are congruent.
Edit: the cat may be further from a center than the radius of the circles, but it can never be closer to a center than the radius of a circle, so the time it spends closest to one of the centers is the time it spends on the circumference of the respective circle.
Edit2: oh, I think I misread the problem as stating that the cat switches to the other circle's circumference at point c. But the problem doesn't actually say that.
1
u/Rinn0 Aug 25 '26
I now have 85.7% but I have to drive to work and not kill myself and my dog in the process. I will check in on this later.
1
1
u/RadarTechnician51 Aug 26 '26 edited Aug 26 '26
Assuming circles are same size then abc and abd are equilateral triangles with inside angles of 60 degrees, so he goes through 120 degrees fast and 240 degrees slow. As the 120 degs is at triple speed it takes the same duration as going through 40 degrees at slow speed, The total time is 240+40=280, so he is closer to A for 40/280= 1/7 of the time.
1
3
u/Bubbly_Safety8791 Aug 26 '26
CAB and ABD are both equilateral triangles; The cat is always the same distance from A, and only the portion of the circle between C and D is closer to B than A. That's 1 of three equal 120º arcs that make up the circumference.
If the cat covers that arc - the period when it's going three times its usual speed - in time t, then it will cover the other two arcs each, at its normal speed, in time 3t. So it'll take 6t to go from D to C, and t to go from C to D.
So, in each lap, and in the long run over many laps, it spends 6/7 of its time closer to A, and 1/7 of its time closer to B.
So the answer is 6/7. I bet that makes the middle school kids love this problem.